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Hans Raj 8 years, 4 months ago
{tex} \sqrt{2}{/tex} = 1.414
{tex} \sqrt{3}{/tex} = 1.732
there are many rational numbers between them , one of them is 3/2 = 1.5
there are many irrational numbers betwen them , one of them
is ({tex}\sqrt{2}{/tex} + {tex} \sqrt{3}{/tex} )/ 2 = 1.573
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Hans Raj 8 years, 4 months ago
let us take an example and solve it by using 8 simple steps
Factor x2 + 5x + 6
1 compare it with standard form ax2 + bx + c
2 a= 1 , c = 6 , axc = 1x6=6
3 next we need to find factors of 6 which add up to 5 ( i.e = b)
4 since 6 can be written as product of 3 and 2 and since 2 + 3 = 5
5 Therefore we will use 2 and 3
6 we will write two numbers found above as follows
7 (x + 2)(x +3)
8 Therefore the answer is x2 + 5x + 6 = (x + 2)(x + 3)
Posted by Ravi Singh 8 years, 4 months ago
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Dharmendra Kumar 8 years, 4 months ago
A number which can be expressed in the form of {tex}p\over q{/tex} (q cannot be 0) , is called rational number.
For example:- 1, 2,{tex}3\over 2{/tex},{tex}-1\over 7{/tex}, {tex}2\over 5{/tex} etc.
A number which cannot be expressed in the form of {tex}p\over q{/tex}
is called an irrational number.
For example:- {tex}{\sqrt 2}{/tex} ,{tex}{\sqrt 5}{/tex} , etc.
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Dharmendra Kumar 8 years, 4 months ago
Heron's formula is used to find the Area of triangle.
Let ABC is triangle and length of sides of triangle ABC are a,b and c.
Then
area of triangleABC ={tex}{\sqrt {s(s-a)(s-b)(s-c)}}{/tex} [HERON'S FORMULA]
where s is a Semi-perimetre of triangle ABC i.e. s={tex}a+b+c\over 2{/tex}
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Hans Raj 8 years, 4 months ago
- {tex} \sqrt{3}{/tex} {tex} /2{/tex}
or
if root under (three by two)
- {tex} \sqrt{3 \over 2}{/tex}
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